(Perhaps, there is a very general answer: not just for k[G] but for an arbitrary algebra — or even arbitrary operad, maybe. Probably, Tyler Lawson's comment is relevant — if somebody could elaborate on that...)

Regarding your edit: If you're comfortable with $A_\infty$ structures, the action of G on K is recoverable from either an $A_\infty$-action of k[G] on K, or an $A_\infty$-map from k[G] to End(K).
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Tyler LawsonApr 18 '11 at 15:01

@Tyler Lawson Oh. You're definitely right (I'd only say "...on H(K)"; if somebody needs details, see Keller's "Introduction to A-infinity algebras and modules"). One thing I still don't understand is how this approach explains the (co)action of $H^2(G)$ from the example (looks not unlike some manifestation of Koszul duality, maybe). But probably I should think about all this a little more before asking new questions.
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Grigory MApr 18 '11 at 17:20

@Grigory: I don't tend to say "on H(K)" because I'm not usually working over a field, but if you are then there's certainly no problem with working that way.
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Tyler LawsonApr 24 '11 at 12:50

3 Answers
3

One family of secondary operations that arise come from trying to find the difference between "classes that look like they are acted on trivially" and "classes that are genuinely (or coherently) acted on trivially".

The spectral sequence starts with
$$E_1^{p,q} = Hom(F_p, H^q(K))$$
and has next term
$$E_2^{p,q} = H^p(G, H^q(K)).$$
In particular, $E_2^{0,q}$ consists of the elements in $H_q(K)$ which are fixed by the $G$-action.

From here, we get a "secondary" operation: the $d_2$-differential $H^0(G, H^q(K)) \to H^2(G, H^{q-1}(K))$. For cyclic groups, you've described this on elements (modulo you need to be careful about how well-defined it is). The higher differentials give tertiary and higher operations. For a given element $x \in H^q(K)$, these measure a sequence of obstructions to finding a chain complex $L \sim K$ such that $x$ is represented in $L$ by a class honestly fixed by $G$.

(I cohomologically indexed $K$, but only because people tend to complain if I don't.)

The spectral sequence reminds me of the hypercohomology spectral sequence. The cases I'm aware of require both complexes to be bounded below or one complex to be bounded below and above in order to ensure convergence. Are there any such restrictions for the current spectral sequence ?
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RalphApr 18 '11 at 21:02

@Ralph: Yes, I am being too casual about this. I would call this a conditionally convergent spectral sequence; it's computing the hyper-Ext of Z into K over Z[G]. If you have boundedness such as you describe, then you are guaranteed that things work out as nicely as possible.
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Tyler LawsonApr 19 '11 at 23:10

Let's assume it is a chain complex of vector spaces over some field. Write the action as $g\mapsto a(g)$ where $a:K\to K$ is a chain map. For each $g\in G$, $a(g)$ is chain homotopic to the identity map. Choose $b(g)$ such that $db(g)+b(g)d=a(g)-1$. Then $db(g)a(h)+b(g)a(h)d=a(gh)-a(h)$ and $da(g)b(h)+a(g)b(h)d=a(gh)-a(g)$, so if $c(g,h)=b(g)a(h)-a(g)b(h)-b(g)+b(h)$ then $dc(g,h)+c(g,h)d=0$. This gives an element of $Hom(H_nK,H_{n+1}K)$ for every $g$ and $h$, and I believe a well-defined element of $H^2(G;Hom(H_nK,H_{n+1}K))$, for every $n$.

Edit: This is the same sort of thing that Tyler got in his answer. I was thinking about it like this: Imagine that it makes sense to speak of the topological group of automorphisms of $K$. We have a map of $G$ into $ker(Aut(K)\to \pi_0(Aut(K))=Aut(HK))$ and thus a map from $BG$ into the classifying space of the latter. This classifying space is simply connected and has $\pi_2=\pi_1Aut(K)=\pi_1End(K)=\prod_n Hom(H_nK,H_{n+1}K)$

It's kind of the same thing, but I was only working out obstructions element-by-element whereas your approach leads to obstructions to trivializing the action on the whole complex at a time - this uses more of the data.
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Tyler LawsonApr 17 '11 at 15:00

1

Over a field, the first thing I see is the first thing you see. Over $\mathbb Z$ I seem to be also getting an element of $H^1(G;Ext(H_nK),H_{n+1}K))$ for all $n$.
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Tom GoodwillieApr 17 '11 at 15:51

1

Well, sure. So in general you should be classifying the (derived) maps of DGAs from Z[G] to Hom(K,K), and so you should get obstructions/classifying elements in Andre-Quillen cohomology $AQ^s(Z[G], H_t Hom(K,K))$, which away from the bottom are $H^{s+1}(G, Ext^{-t}(K,K))$. Specializing to what you get on homology groups should be producing the elements you describe.
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Tyler LawsonApr 17 '11 at 16:40

(Nothing really new here, but just to link $A_\infty$-description and the explicit construction.)

Action of G on K induces $A_\infty$-action of Z[G] on H(K)=:H. And this is all operations one can define on H(K), since this structure defines K up to (equivariant) q/iso (ref).

Now, let's rewrite this in more concrete terms. $A_\infty$-action of Z[G] on H is just a collection of maps $m_i\colon G^{i-1}\times H\to H$ aka $m_i\colon G^{i-1}\to\operatorname{Hom}(H,H)$ ($i=2,3,\dots$). This maps are subject to some relations. In particular, if $m_{i+1}$ is first non-zero higher action, then $m_{i+1}$ satisfies cocycle condition and gives an element in $H^i(G,\operatorname{Hom}(H,H))$. For i=2 it is exactly the element from Tom Goodwillie's answer (and, probably, it also coincides with $d_i$ from Tyler Lawson's answer, although I don't have a proof).